Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class Major{
    static int array[][];
    static int posx, posy;
    static void det(){
        int min = 600;
        for(int i = 0;i < array.length;i++){
            for(int j = 0;j < array[0].length;j++){
                if(array[i][j] <= 0)continue;
                if(min > array[i][j]){
                    min = array[i][j];
                    posx = i;
                    posy = j;
                }
            }
        }
    }
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());
        array = new int[n][m];
        for(int i = 0;i < n;i++){
            st = new StringTokenizer(br.readLine());
            for(int j = 0;j < m;j++){
                array[i][j] = Integer.parseInt(st.nextToken());
            }
        }
        int ans = 0;
        StringBuilder sb = new StringBuilder();
        if(n <= m) {
            while (true) {
                posx = -1;
                posy = -1;
                det();
                if (posx == -1) break;
                boolean flag = true;
                for (int i = 0; i < m; i++)
                    if (array[posx][i] <= 0) flag = false;
                ans += array[posx][posy];
                int x = array[posx][posy];
                if (flag) {
                    for (int i = 0; i < m; i++) {
                        array[posx][i] -= x;
                    }
                    for (int i = 0; i < x; i++) {
                        sb.append("row " + (posx + 1));
                        sb.append('\n');
                    }
                } else {
                    for (int i = 0; i < n; i++) {
                        array[i][posy] -= x;
                    }
                    for (int i = 0; i < x; i++) {
                        sb.append("col " + (posy + 1));
                        sb.append('\n');
                    }
                }
            }
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    if (array[i][j] != 0) {
                        System.out.println("-1");
                        return;
                    }
                }
            }
            System.out.println(ans);
            System.out.println(sb);
        }
        else {
            while(true){
                posx = -1;
                posy = -1;
                det();
                if(posx == -1)break;
                boolean flag = true;
                for(int i = 0;i < n;i++)
                    if(array[i][posy] <= 0)flag = false;
                ans += array[posx][posy];
                int x = array[posx][posy];
                if(flag) {
                    for (int i = 0; i < n; i++) {
                        array[i][posy] -= x;
                    }
                    for (int i = 0; i < x; i++){
                        sb.append("col " + (posy + 1));
                        sb.append('\n');
                    }
                }
                else {
                    for (int i = 0; i < m; i++) {
                        array[posx][i] -= x;
                    }
                    for (int i = 0; i < x; i++){
                        sb.append("row " + (posx + 1));
                        sb.append('\n');
                    }
                }
            }
            for(int i = 0;i < n;i++) {
                for (int j = 0; j < m; j++) {
                    if (array[i][j] != 0) {
                        System.out.println("-1");
                        return;
                    }
                }
            }
            System.out.println(ans);
            System.out.println(sb);
        }
    }
}